Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2010 AIME I Problems/Problem 10"

(Remove extra problem section)
m (→‎Solution 5)
 
(13 intermediate revisions by 8 users not shown)
Line 6: Line 6:
  
 
== Solution 2 ==
 
== Solution 2 ==
−
Note that <math>a_2\cdot 10^2 + a_0</math> is the base <math>100</math> representation of any number from <math>0</math> to <math>9999</math>, and similarly <math>10(a^3\cdot 10^2 + a_1)</math> is ten times the base <math>100</math> representation of any number from <math>0</math> to <math>9999</math>. Thus, the number of solution is just the number of solutions to <math>2010 = 10a+b</math> where <math>0\le a, b\le 9999</math>, which is clearly equal to <math>\boxed{202}</math> as <math>a</math> can range from <math>0</math> to <math>201</math>.
+
Note that <math>a_2\cdot 10^2 + a_0</math> is the base <math>100</math> representation of any number from <math>0</math> to <math>9999</math>, and similarly <math>10(a_3\cdot 10^2 + a_1)</math> is ten times the base <math>100</math> representation of any number from <math>0</math> to <math>9999</math>. Thus, the number of solutions is just the number of solutions to <math>2010 = 10a+b</math> where <math>0\le a, b\le 9999</math>, which is equal to <math>\boxed{202}</math> as <math>a</math> can range from <math>0</math> to <math>201</math>.
  
 
== Solution 3 ==
 
== Solution 3 ==
Line 22: Line 22:
  
 
Case 1: <math>a_3 = 0</math>.
 
Case 1: <math>a_3 = 0</math>.
−
We try to find possible values of <math>a_2</math>. We plug in <math>a_3 = 0</math> and <math>10a_1 + a_0 = 1089</math> to our initial equation, which gives us <math>2010 = 0 + 100a_2^2 + 1089</math>. Thus <math>a_2 \geq 10</math>. We also see that <math>a_2 \leq 20</math>. We now take these values of <math>a_2</math> and find the number of pairs <math>(a_1, a_0)</math> that work. If <math>a_2 = 10</math>, <math>10a_1 + a_0 = 1010</math>. We see that there are <math>8</math> possible pairs in this case. Using the same logic, there are <math>10</math> ways for <math>a_2 = 11, 12 \ldots 19</math>. For <math>a_2 = 20</math>, we get the equation <math>10a_1 + a_0 = 10</math>, for 2 ways. Thus, for <math>a_3 = 0</math>, there are <math>8 + 10 \cdot 9 + 2 = 100</math> ways.
+
We try to find possible values of <math>a_2</math>. We plug in <math>a_3 = 0</math> and <math>10a_1 + a_0 = 1089</math> to our initial equation, which gives us <math>2010 = 0 + 100a_2 + 1089</math>. Thus <math>a_2 \geq 10</math>. We also see that <math>a_2 \leq 20</math>. We now take these values of <math>a_2</math> and find the number of pairs <math>(a_1, a_0)</math> that work. If <math>a_2 = 10</math>, <math>10a_1 + a_0 = 1010</math>. We see that there are <math>8</math> possible pairs in this case. Using the same logic, there are <math>10</math> ways for <math>a_2 = 11, 12 \ldots 19</math>. For <math>a_2 = 20</math>, we get the equation <math>10a_1 + a_0 = 10</math>, for 2 ways. Thus, for <math>a_3 = 0</math>, there are <math>8 + 10 \cdot 9 + 2 = 100</math> ways.
  
 
Case 2: <math>a_3 = 1</math>.
 
Case 2: <math>a_3 = 1</math>.
Line 31: Line 31:
  
 
Totaling everything, we get that there are <math>100 + 100 + 2 = \boxed{202}</math> ways.
 
Totaling everything, we get that there are <math>100 + 100 + 2 = \boxed{202}</math> ways.
 +
 +
==Solution 5: Generating Functions==
 +
We will represent the problem using generating functions. Consider the generating function <cmath>f(x) = (1+x^{1000}+x^{2000}+\cdots+x^{99000})(1+x^{100}+x^{200}+\cdots+x^{9900})(1+x^{10}+x^{20}+\cdots+x^{990})(1+x+x^2+\cdots+x^{99})</cmath>
 +
where the first factor represents <math>a_3</math>, the second factor <math>a_2</math>, and so forth. We want to find the coefficient of <math>x^{2010}</math> in the expansion of <math>f(x)</math>. Now rewriting each factor using the geometric series yields <cmath>f(x) = \frac{\cancel{x^{100}-1}}{x-1} \cdot \frac{\cancel{x^{1000}-1}}{x^{10}-1} \cdot \frac{x^{10000}-1}{\cancel{x^{100}-1}} \cdot \frac{x^{100000}-1}{\cancel{x^{1000}-1}}=\frac{x^{10000}-1}{x-1} \cdot \frac{x^{100000}-1}{x^{10}-1} = (1+x+x^2+\cdots + x^{9999})(1+x^{10}+x^{20}+\cdots+x^{99990})</cmath>
 +
The coefficient of <math>x^{2010}</math> in this is simply <math>\boxed{202}</math>, as we can choose any of the first 202 terms from the second factor and pair it with exactly one term in the first factor.
 +
 +
~rzlng
 +
==Solution 5==
 +
First note that <math>a_3</math> has to be a single-digit number(<math>0</math>, <math>1</math>, or <math>2</math> to be exact), and that <math>a_1</math> has to be a two-digit multiple of ten.
 +
Then, <math>a_3</math>, <math>a_2</math>, <math>a_1</math> and <math>a_0</math> can be represented as follows:
 +
<cmath>\begin{align*}
 +
a_3 = a
 +
\\ a_2 = 10b+c
 +
\\ a_1= 10d+e
 +
\\ a_0 = 10f
 +
\end{align*}</cmath>
 +
, where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math>, and <math>f</math> are all(not necessarily nonzero) digits.
 +
Now, we can write our given equation as follows:
 +
<cmath>
 +
\begin{align*}
 +
2010 = 1000(a+b) + 100(c+d) + 10(e+f)
 +
\\ 201 = 100(a+b) + 10(c+d) + (e+f)
 +
\\ 201 = (100a + 10c + e) + (100b + d + f)
 +
\end{align*}
 +
</cmath>
 +
Now, each integer between <math>0</math> and <math>201</math> inclusive can be represented in exactly one way as <math>100a + 10c + e</math>, and this corresponds with one unique <math>100b + d + f</math>, so it remains to count the number of integers between <math>0</math> and <math>201</math> inclusive. This is easily counted to be <math>\boxed{202}</math>.
 +
 +
==Solution 6 ==
 +
Just note that this corresponds to <math>0 \leq 10^3 \cdot a_3 + 10^2 \cdot a_2 + 10^1 \cdot a_1 \leq 2010</math>, because we can use <math>a_0</math> to fill in the remaining gap. Then, dividing by <math>10</math>, we have <math>0 \leq \overline{a_3 a_2 a_1} \leq 201</math>, of which there are <math>\boxed{202}</math> solutions.
 +
 +
==Video Solution==
 +
 +
https://youtu.be/DNC2zd5rReY
  
 
== See Also ==
 
== See Also ==

Latest revision as of 07:07, 15 August 2024

Problem

Let $N$ be the number of ways to write $2010$ in the form $2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$, where the $a_i$'s are integers, and $0 \le a_i \le 99$. An example of such a representation is $1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0$. Find $N$.

Solution 1

If we choose $a_3$ and $a_1$ such that $(10^3)(a_3) + (10)(a_1) \leq 2010$ there is a unique choice of $a_2$ and $a_0$ that makes the equality hold. So $N$ is just the number of combinations of $a_3$ and $a_1$ we can pick. If $a_3 = 0$ or $a_3 = 1$ we can let $a_1$ be anything from $0$ to $99$. If $a_3 = 2$ then $a_1 = 0$ or $a_1 = 1$. Thus $N = 100 + 100 + 2 = \fbox{202}$.

Solution 2

Note that $a_2\cdot 10^2 + a_0$ is the base $100$ representation of any number from $0$ to $9999$, and similarly $10(a_3\cdot 10^2 + a_1)$ is ten times the base $100$ representation of any number from $0$ to $9999$. Thus, the number of solutions is just the number of solutions to $2010 = 10a+b$ where $0\le a, b\le 9999$, which is equal to $\boxed{202}$ as $a$ can range from $0$ to $201$.

Solution 3

Note that $a_0 \equiv 2010\ (\textrm{mod}\ 10)$ and $a_1 \equiv 2010 - a_0\ (\textrm{mod}\ 100)$. It's easy to see that exactly 10 values in $0 \leq a_0 \leq 99$ that satisfy our first congruence. Similarly, there are 10 possible values of $a_1$ for each choice of $a_0$. Thus, there are $10 \times 10 = 100$ possible choices for $a_0$ and $a_1$. We next note that if $a_0$ and $a_1$ are chosen, then a valid value of $a_3$ determines $a_2$, so we dive into some simple casework:

  • If $2010 - 10a_1 - a_0 \geq 2000$, there are 3 valid choices for $a_3$. There are only 2 possible cases where $2010 - 10a_1 - a_0 \geq 2000$, namely $(a_1, a_0) = (1,0), (10,0)$. Thus, there are $3 \times 2 = 6$ possible representations in this case.
  • If $2010 - 10a_1 - a_0 < 1000$, $a_3$ can only equal 0. However, this case cannot occur, as $10a_1+a_0\leq 990+99 = 1089$. Thus, $2010-10a_1-a_0 \geq 921$. However, $2010-10a_1-a_0 = 1000a_3 + 100a_2 \equiv 0\ (\textrm{mod}\ 100)$. Thus, we have $2010-10a_1-a_0 \geq 1000$ always.
  • If $1000 \leq 2010 - 10a_1 - a_0 < 2000$, then there are 2 valid choices for $a_3$. Since there are 100 possible choices for $a_0$ and $a_1$, and we have already checked the other cases, it follows that $100 - 2 - 0 = 98$ choices of $a_0$ and $a_1$ fall under this case. Thus, there are $2 \times 98 = 196$ possible representations in this case.

Our answer is thus $6 + 0 + 196 = \boxed{202}$.

Solution 4: Casework and Brute Force

We immediately see that $a_3$ can only be $0$, $1$ or $2$. We also note that the maximum possible value for $10a_1 + a_0$ is $990 + 99 = 1089$. We then split into cases:

Case 1: $a_3 = 0$. We try to find possible values of $a_2$. We plug in $a_3 = 0$ and $10a_1 + a_0 = 1089$ to our initial equation, which gives us $2010 = 0 + 100a_2 + 1089$. Thus $a_2 \geq 10$. We also see that $a_2 \leq 20$. We now take these values of $a_2$ and find the number of pairs $(a_1, a_0)$ that work. If $a_2 = 10$, $10a_1 + a_0 = 1010$. We see that there are $8$ possible pairs in this case. Using the same logic, there are $10$ ways for $a_2 = 11, 12 \ldots 19$. For $a_2 = 20$, we get the equation $10a_1 + a_0 = 10$, for 2 ways. Thus, for $a_3 = 0$, there are $8 + 10 \cdot 9 + 2 = 100$ ways.

Case 2: $a_3 = 1$. This case is almost identical to the one above, except $0 \leq a_2 \leq 10$. We also get 100 ways.

Case 3: $a_3 = 2$. If $a_3 = 2$, our initial equation becomes $100a_2 + 10a_1 + a_0 = 10$. It is obvious that $a_2 = 0$, and we are left with $10a_1 + a_0 = 10$. We saw above that there are $2$ ways.

Totaling everything, we get that there are $100 + 100 + 2 = \boxed{202}$ ways.

Solution 5: Generating Functions

We will represent the problem using generating functions. Consider the generating function \[f(x) = (1+x^{1000}+x^{2000}+\cdots+x^{99000})(1+x^{100}+x^{200}+\cdots+x^{9900})(1+x^{10}+x^{20}+\cdots+x^{990})(1+x+x^2+\cdots+x^{99})\] where the first factor represents $a_3$, the second factor $a_2$, and so forth. We want to find the coefficient of $x^{2010}$ in the expansion of $f(x)$. Now rewriting each factor using the geometric series yields \[f(x) = \frac{\cancel{x^{100}-1}}{x-1} \cdot \frac{\cancel{x^{1000}-1}}{x^{10}-1} \cdot \frac{x^{10000}-1}{\cancel{x^{100}-1}} \cdot \frac{x^{100000}-1}{\cancel{x^{1000}-1}}=\frac{x^{10000}-1}{x-1} \cdot \frac{x^{100000}-1}{x^{10}-1} = (1+x+x^2+\cdots + x^{9999})(1+x^{10}+x^{20}+\cdots+x^{99990})\] The coefficient of $x^{2010}$ in this is simply $\boxed{202}$, as we can choose any of the first 202 terms from the second factor and pair it with exactly one term in the first factor.

~rzlng

Solution 5

First note that $a_3$ has to be a single-digit number($0$, $1$, or $2$ to be exact), and that $a_1$ has to be a two-digit multiple of ten. Then, $a_3$, $a_2$, $a_1$ and $a_0$ can be represented as follows: \begin{align*} a_3 = a \\ a_2 = 10b+c \\ a_1= 10d+e \\ a_0 = 10f \end{align*} , where $a$, $b$, $c$, $d$, $e$, and $f$ are all(not necessarily nonzero) digits. Now, we can write our given equation as follows: \begin{align*} 2010 = 1000(a+b) + 100(c+d) + 10(e+f) \\ 201 = 100(a+b) + 10(c+d) + (e+f) \\ 201 = (100a + 10c + e) + (100b + d + f) \end{align*} Now, each integer between $0$ and $201$ inclusive can be represented in exactly one way as $100a + 10c + e$, and this corresponds with one unique $100b + d + f$, so it remains to count the number of integers between $0$ and $201$ inclusive. This is easily counted to be $\boxed{202}$.

Solution 6

Just note that this corresponds to $0 \leq 10^3 \cdot a_3 + 10^2 \cdot a_2 + 10^1 \cdot a_1 \leq 2010$, because we can use $a_0$ to fill in the remaining gap. Then, dividing by $10$, we have $0 \leq \overline{a_3 a_2 a_1} \leq 201$, of which there are $\boxed{202}$ solutions.

Video Solution

https://youtu.be/DNC2zd5rReY

See Also

2010 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_10&oldid=225049"